What is the reduced row echelon form of a matrix

To convert a matrix to the reduced row-echelon form, we ensure the following conditions:

  • Each leading entry (first non-zero element) for a non-zero row should be equal to 1 and should have all zero elements to its left, bottom, and above.

  • All zero rows should be placed at the end of the matrix.

For a matrix A, we represent this as:

A=[100013000]A = \begin{bmatrix}1 & 0 &0 \\0 & 1 & 3 \\0 & 0 & 0\\\end{bmatrix}

Here the leading entry for the 1st row and 2nd row is11, and the elements below and to the left are 0. The zero row is placed at the end.

Basic and non-basic column

In a matrix of reduced row echelon form, basic columns are matrix columns that contain a leading entry, while non-basic columns are columns that don't have a leading entry.

Following is the example of a matrix AA:

A=[1200]A = \begin{bmatrix}1 & 2 \\0 & 0 \\\end{bmatrix}

Here the 1st column is a basic column with leading entry 11 while the 2nd column is a non-basic column with no leading entry.

Converting a matrix to reduced row-echelon form

To convert a simple matrix to reduced row echelon form, we use the Gaussian elimination method approach. This approach mainly involves the following steps:

  • Interchanging any two rows

  • Adding any two rows

  • Multiplying a row by a scalar with a non-zero value.

Note: We can perform any two of the above steps together at the same time.

Example

Consider an example of converting a matrix to a reduced row-echelon form. Suppose we have a matrix AA such that:

A=[123356123]A = \begin{bmatrix}1 & 2 &3 \\3 & 5 & 6 \\1 & 2 & 3\\\end{bmatrix}

  • The leading entry of the 1st row is 1. Since it is the 1st row, there are no elements above it, so we only convert all the elements below it to 0.

We multiply the 1st row by -3 and add it to the 2nd row:

A=[1233+(−3×1)5+(−3×2)6+(−3×3)123]A = \begin{bmatrix}1 &2 &3 \\3+(-3×1) & 5 +(-3×2) & 6+(-3×3) \\1 & 2 & 3\\\end{bmatrix}
A=[1230−1−3123]A = \begin{bmatrix}1 & 2 &3 \\0 & -1& -3\\1 & 2 & 3\\\end{bmatrix}

We multiply the 1st row by -1 and add it to the 3rd row:

A=[1230−1−31+(−1×1)2+(−1×2)3+(−1×3)]A = \begin{bmatrix}1 & 2 &3 \\0 & -1 & -3\\1+(-1×1) & 2+(-1×2) & 3+(-1×3)\\\end{bmatrix}
A=[1230−1−3000]A = \begin{bmatrix}1 & 2 &3 \\0 & -1 & -3\\0 & 0 & 0\\\end{bmatrix}

  • The leading entry of the 2nd row is -1, so we multiply the 2nd row with -1.

A=[1230×−1−1×−1−3×−1000]A = \begin{bmatrix}1 & 2 &3 \\0 ×-1& -1 ×-1& -3×-1\\0 & 0 & 0\\\end{bmatrix}
A=[123013000]A = \begin{bmatrix}1 & 2 &3 \\0 & 1 & 3\\0 & 0 & 0\\\end{bmatrix}

  • The leading entry of the 2nd row is now 1, and all the elements below it are already 0. We convert the element above it to 0 now.

A=[123013000]A = \begin{bmatrix}1 & 2 &3 \\0 & 1 & 3\\0 & 0 & 0\\\end{bmatrix}

We multiply the 2nd row by -2 and add it to the 1st row:

A=[1+(−2×0)2+(−2×1)3+(−2×3)013000]A = \begin{bmatrix}1+(-2×0) & 2 +(-2×1)&3+(-2×3) \\0 & 1 & 3\\0 & 0 & 0\\\end{bmatrix}
A=[10−3013000]A = \begin{bmatrix}1 & 0 &-3 \\0 & 1 & 3\\0 & 0 & 0\\\end{bmatrix}

Our final result is:

A=[123013000]A = \begin{bmatrix}1 & 2 &3 \\0 & 1 & 3\\0 & 0 & 0\\\end{bmatrix}

Conclusion

The reduced row echelon form is a powerful tool in linear algebra. It helps us simplify our matrices so we can apply operations on them. Moreover, it gives a clear structure to determine relationships within matrices, contributing to efficient problem-solving across various fields.

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